Since the question is for dh/dt when h=4ft at the 9-ft deep end, then the volume of a trapezoidal prism won't work. Because at h=4ft, you only have a triangular prism for the volume of water in the tank.
Draw the figure, a longitudinal section of the pool, such that the walls are vertical. It is a trapezoid, (trapezium).
Draw horizontal line passing through the 4-ft deep.
Draw another horizontal line lower than tha first line.
Two similar right triangles are formed.
----vertical leg = (9-4) = 5ft
----horizontal leg = 40ft
The smaller one has:
---vertical leg = h
---horizontal leg = x
5/40 = h/x
5x = 40h
x = 40h/5 = 8h
So, volume of water in the pool when depth at deep end is h ft,
V = (1/2)(h)(x) *(20)
V = 10h(8h)
V = 80h^2
dV/dt = 80(2h *dh/dt)
dV/dt = 160h (dh/dt)
When h=4, and dV/dt is always 10 cuft/min,
10 = 160(4)(dh/dt)
1 = 64(dh/dt)
dh/dt = 1/64
Therefore, at that instant, the water is rising at the rate of 1/64 ft/min. ----answer.